Question

How much work does the charge escalator do to move $1.0\mu C$ of charge from the negative terminal to the positive terminal of a $1.5V$ battery?

Short answer

The work done is$\mathbf{1}\mathbf{.}\mathbf{5}\mathit{\mu }\mathbf{}\mathit{J}$

Step-by-step solution

Given information and theory used  

Given :

Charge :$1.0\mu C$

Battery Voltage : $1.5V$

Theory used :

The work is equal to$W=q∆V$

Calculating Work done

The work is equal to the charge's change in potential energy, which is equal to the charge multiplied by the potential difference: $W=q∆V$

$$\begin{gathered} \Rightarrow W=q∆V \\ =1.0\mu C·1.5V \\ =\boxed{\mathbf{1}\mathbf{.}\mathbf{5}\mathit{\mu }\mathbf{}\mathit{J}} \end{gathered}$$